Simultaneous linear & quadratic equations

You should compare the coefficients in the two equations, and look for like terms: if you collect those terms, which terms will cancel?

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is elimination. The elimination method works by cancelling like terms which the two equations share. We need to figure out which of the terms in the equations will cancel, and why that cancellation happens.

In these two questions, there are two $x$-terms and two $y$-terms:

The goal of elimination is to cancel out like terms by adding (or subtracting) the equations. That can be either the $x$-terms or the $y$-terms.

For the equations in this question, notice that the coefficients of $x$ are equal and opposite. So if we add them together, they will cancel. So we will add the equations together to make that cancellation happen, like this:

The $x$-terms cancel each other out - that is what 'elimination' means! We get the $0x$ because the $x$-coefficients are equal and opposite: $4x + \left(-4x\right)$ is equal to zero. So the elimination depends on which of the coefficients cancel out.

The correct answer choices are: We will eliminate the $x$-terms because the $x$-coefficients will cancel.

You should compare the coefficients in the two equations, and look for like terms: if you collect those terms, which terms will cancel?

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is elimination. The elimination method works by cancelling like terms which the two equations share. We need to figure out which of the terms in the equations will cancel, and why that cancellation happens.

In these two questions, there are two $x$-terms and two $y$-terms:

The goal of elimination is to cancel out like terms by adding (or subtracting) the equations. That can be either the $x$-terms or the $y$-terms.

For the equations in this question, notice that the coefficients of $x$ are equal and opposite. So if we add them together, they will cancel. So we will add the equations together to make that cancellation happen, like this:

The $x$-terms cancel each other out - that is what 'elimination' means! We get the $0x$ because the $x$-coefficients are equal and opposite: $-5x + \left(5x\right)$ is equal to zero. So the elimination depends on which of the coefficients cancel out.

The correct answer choices are: We will eliminate the $x$-terms because the $x$-coefficients will cancel.

You should compare the coefficients in the two equations, and look for like terms: if you collect those terms, which terms will cancel?

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is elimination. The elimination method works by cancelling like terms which the two equations share. We need to figure out which of the terms in the equations will cancel, and why that cancellation happens.

In these two questions, there are two $x$-terms and two $y$-terms:

The goal of elimination is to cancel out like terms by adding (or subtracting) the equations. That can be either the $x$-terms or the $y$-terms.

For the equations in this question, notice that the coefficients of $x$ are equal and opposite. So if we add them together, they will cancel. So we will add the equations together to make that cancellation happen, like this:

The $x$-terms cancel each other out - that is what 'elimination' means! We get the $0x$ because the $x$-coefficients are equal and opposite: $2x + \left(-2x\right)$ is equal to zero. So the elimination depends on which of the coefficients cancel out.

The correct answer choices are: We will eliminate the $x$-terms because the $x$-coefficients will cancel.

The first thing to do is compare the coefficients of the terms. We can see that the coefficient of the $n$-term in the first equation is a multiple of the second $n$-term: the $\text{2}$ is $\text{5}$ times as much as $\text{10}$. If we multiply the entire second equation by this factor of $\text{5}$ we will be able to cancel the $n$-terms between the equations:

Now eliminate the $n$-terms. Since the terms have opposite signs, we must add one equation to the other to cancel those terms.

Now we can solve for $m$:

Finally, use the value of $m$ to find the value of $n$. Remember that we can use either of the equations to do this, we can pick the easier equation. The second equation looks like a better choice because it has fewer negative signs than the first equation (in fact it does not have any!).

The answers to the pair of equations are $m = 3$ and $n = 4$.

The first thing to do is compare the coefficients of the terms. We can see that the coefficient of the $j$-term in the first equation is a multiple of the second $j$-term: the $\text{5}$ is $\text{5}$ times as much as $\text{25}$. If we multiply the entire second equation by this factor of $\text{5}$ we will be able to cancel the $j$-terms between the equations:

Now eliminate the $j$-terms. Since the terms have the same signs, we must subtract one equation from the other to cancel those terms.

Now we can solve for $k$:

Finally, use the value of $k$ to find the value of $j$. Remember that we can use either of the equations to do this, we can pick the easier equation. The equations are pretty much the same - for example, neither has any fractions or negatives. So we will use the first one.

The answers to the pair of equations are $j = 1$ and $k = -1$.

The first thing to do is compare the coefficients of the terms. We can see that the coefficient of the $n$-term in the first equation is a multiple of the second $n$-term: the $\text{4}$ is $\text{4}$ times as much as $\text{16}$. If we multiply the entire second equation by this factor of $\text{4}$ we will be able to cancel the $n$-terms between the equations:

Now eliminate the $n$-terms. Since the terms have opposite signs, we must add one equation to the other to cancel those terms.

Now we can solve for $m$:

Finally, use the value of $m$ to find the value of $n$. Remember that we can use either of the equations to do this, we can pick the easier equation. In this case the first equation has fewer negatives than the second equation. So we will use the first equation.

The answers to the pair of equations are $m = -5$ and $n = 1$.

Notice that the $y$-coefficients are opposites. You should start by adding the equations together, which will cancel those terms out.

We have two equations and we need to solve them 'simultaneously'. That means we need to find two numbers, for $x$ and for $y$, which solve both of the equations.

There are two methods for solving equations simultaneously: elimination and substitution. The equations in this question are perfect for elimination because the $y$-coefficients are equal and opposite. If we add the equations together, the $y$-terms will cancel. (We could solve these equations using the substitution method but elimination will be faster.)

So we add the equations together. This means we add together the left sides of the equations and also the right sides of the equations.

Now we have an equation with only one variable, and we can solve it.

So $x = -2$. But we are not done yet: we also need an answer for $y$. We can get this using the answer we just got for $x$. Substitute this into either of the equations to get the answer - we can use either equation because we are looking for the number that solves both of them. We will pick the second equation, because it has fewer negative signs.

Now we have the complete answer: the numbers which solve the equations are $x = -2$ and $y = -1$. We can see this represented beautifully on a graph of the two equations: the answers we just found are the coordinates of the point where the lines intersect.

The values which solve these two equations are $x = -2$ and $y = -1$.

Notice that the $x$-coefficients are opposites. You should start by adding the equations together, which will cancel those terms out.

We have two equations and we need to solve them 'simultaneously'. That means we need to find two numbers, for $x$ and for $y$, which solve both of the equations.

There are two methods for solving equations simultaneously: elimination and substitution. The equations in this question are perfect for elimination because the $x$-coefficients are equal and opposite. If we add the equations together, the $x$-terms will cancel. (We could solve these equations using the substitution method but elimination will be faster.)

So we add the equations together. This means we add together the left sides of the equations and also the right sides of the equations.

Now we have an equation with only one variable, and we can solve it.

So $y = -2$. But we are not done yet: we also need an answer for $x$. We can get this using the answer we just got for $y$. Substitute this into either of the equations to get the answer - we can use either equation because we are looking for the number that solves both of them. In this case, neither of the equations is much more friendly than the other, so we will just use the first equation.

Now we have the complete answer: the numbers which solve the equations are $x = -3$ and $y = -2$. We can see this represented beautifully on a graph of the two equations: the answers we just found are the coordinates of the point where the lines intersect.

The values which solve these two equations are $x = -3$ and $y = -2$.

Notice that the $y$-coefficients are opposites. You should start by adding the equations together, which will cancel those terms out.

We have two equations and we need to solve them 'simultaneously'. That means we need to find two numbers, for $x$ and for $y$, which solve both of the equations.

There are two methods for solving equations simultaneously: elimination and substitution. The equations in this question are perfect for elimination because the $y$-coefficients are equal and opposite. If we add the equations together, the $y$-terms will cancel. (We could solve these equations using the substitution method but elimination will be faster.)

So we add the equations together. This means we add together the left sides of the equations and also the right sides of the equations.

Now we have an equation with only one variable, and we can solve it.

So $x = -6$. But we are not done yet: we also need an answer for $y$. We can get this using the answer we just got for $x$. Substitute this into either of the equations to get the answer - we can use either equation because we are looking for the number that solves both of them. We will pick the first equation, because it has fewer negative signs.

Now we have the complete answer: the numbers which solve the equations are $x = -6$ and $y = 0$. We can see this represented beautifully on a graph of the two equations: the answers we just found are the coordinates of the point where the lines intersect.

The values which solve these two equations are $x = -6$ and $y = 0$.

You can solve the equations simultaneously using elimination or substitution. The first thing you should do is decide which method is the better choice for these equations.

To solve the given equations simultaneously, we can use either substitution or elimination. The best choice depends on the arrangement of the equations. In this case we need to modify the equations before we can do elimination or substitution. And we only need to make a small change to arrange the equations for substitution: we just need to subtract the $10$ term to the other side of the first equation.

First rearrange the first equation to set up the substitution:

Now do the substitution and complete the solution to find $x$.

Super - we know that $x = -2$.

Now we can use the $x$-value to find the value of $y$. We can do this using either of the equations in the question. If one of the equations is simpler or more convenient (for example, if it has an isolated variable or has fewer negative signs) it is better to choose that equation. In this case, the second equation is better because it has fewer negative signs.

The answer is the pair of numbers $x = -2$ and $y = 0$. On the Cartesian plane, this looks like this:

The values which solve these two equations are $x = -2$ and $y = 0$.

You can solve the equations simultaneously using elimination or substitution. The first thing you should do is decide which method is the better choice for these equations.

To solve the given equations simultaneously, we can use either substitution or elimination. The best choice depends on the arrangement of the equations. These equations are not arranged nicely for elimination or substitution. However, notice that the coefficients in the second equation are all a multiple of $\text{2}$. So we can divide the equation by $\text{2}$ to isolate $x$ without picking up any fractions!

First rearrange the second equation to set up the substitution:

Now do the substitution and complete the solution to find $y$.

Super - we know that $y = 1$.

Now we can use the $y$-value to find the value of $x$. We can do this using either of the equations in the question. If one of the equations is simpler or more convenient (for example, if it has an isolated variable or has fewer negative signs) it is better to choose that equation. In this case, the second equation is better because it has fewer negative signs.

The answer is the pair of numbers $x = 3$ and $y = 1$. On the Cartesian plane, this looks like this:

The values which solve these two equations are $x = 3$ and $y = 1$.

You can solve the equations simultaneously using elimination or substitution. The first thing you should do is decide which method is the better choice for these equations.

To solve the given equations simultaneously, we can use either substitution or elimination. The best choice depends on the arrangement of the equations. These equations invite elimination, because the $x$-terms have equal coefficients. We can eliminate the $x$-terms by subtracting the equations.

Eliminate the $x$ and solve for $y$:

Super - we know that $y = -1$.

Now we can use the $y$-value to find the value of $x$. We can do this using either of the equations in the question. If one of the equations is simpler or more convenient (for example, if it has an isolated variable or has fewer negative signs) it is better to choose that equation. In this case the equations are pretty similar, so we will just pick the first one.

The answer is the pair of numbers $x = -1$ and $y = -1$. On the Cartesian plane, this looks like this:

The values which solve these two equations are $x = -1$ and $y = -1$.

You need to solve these equations using the substitution method. So the first thing to do is pick one of the variables to isolate.

The question tells us to solve the equations using the substitution method. That is not necessarily the best method, but we do not have a choice!

None of the variables are isolated, so we will need to isolate one of them for the substitution step. It would be convenient if there was a coefficient of $\text{1}$ or $-\text{1}$, because that would make our work easier. But that is not the case. So we will just make $x$ the subject of the first equation:

Unfortunately, this equation has $\text{2}$ fractions in it. That is ugly, but it will work.

Substitute the expression for $x$ into the second equation. Then simplify and solve for the value of $y$:

Substitute this value of $y$ back into one of the equations and solve for $x$. In this case, we should use the first equation because it has no negative signs.

The final answers are $x = \dfrac{5}{4}$ and $y = \dfrac{5}{6}$.

You need to solve these equations using the substitution method. So the first thing to do is pick one of the variables to isolate.

The question tells us to solve the equations using the substitution method. That is not necessarily the best method, but we do not have a choice!

None of the variables are isolated, so we will need to isolate one of them for the substitution step. Luckily for us, we can isolate $x$ in the second equation pretty easily because the coefficient of that term is $\text{1}$.

Substitute the expression for $x$ into the first equation. Then simplify and solve for the value of $y$:

Substitute this value of $y$ back into one of the equations and solve for $x$. In this case, the equations are similar: they both have $\text{2}$ negative signs. We will pick the first equation.

The final answers are $x = - \dfrac{19}{7}$ and $y = \dfrac{2}{7}$.

You need to solve these equations using the substitution method. So the first thing to do is pick one of the variables to isolate.

The question tells us to solve the equations using the substitution method. That is not necessarily the best method, but we do not have a choice!

None of the variables are isolated, so we will need to isolate one of them for the substitution step. Luckily for us, we can isolate $x$ in the first equation pretty easily because the coefficient of that term is $\text{1}$.

Substitute the expression for $x$ into the second equation. Then simplify and solve for the value of $y$:

Substitute this value of $y$ back into one of the equations and solve for $x$. In this case, we should use the first equation because it has no negative signs.

The final answers are $x = - \dfrac{37}{13}$ and $y = \dfrac{10}{13}$.

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is $gh$.

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by $gh$, which is the LCD of the fractions. That will cancel all of the denominators.

Now we are ready to solve the following equations:

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by $\text{2}$ so that the $h$-coefficients will be ready to cancel. Then we can eliminate them by subtracting the equations.

First modify things to set up the elimination:

Now do the elimination and complete the solution to find $g$.

Great - we have the first value, $g = 8$.

Now the last step: find the value of $h$. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

The answer is the pair of numbers $g = 8$ and $h = 10$. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

The values which solve these two equations are $g = 8$ and $h = 10$.

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is $ab$.

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by $ab$, which is the LCD of the fractions. That will cancel all of the denominators.

Now we are ready to solve the following equations:

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

In this case we need to modify the equations before we can do elimination or substitution. And we only need to make a small change to arrange the equations for substitution: we just need to multiply both sides of the equation by $-\text{1}$ to remove the negative sign from $a$.

First rearrange things to set up the substitution:

Now do the substitution and complete the solution to find $b$.

Great - we have the first value, $b = - \frac{3}{2}$.

Now the last step: find the value of $a$. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

The answer is the pair of numbers $a = - \frac{1}{2}$ and $b = - \frac{3}{2}$. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

The values which solve these two equations are $a = - \frac{1}{2}$ and $b = - \frac{3}{2}$.

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is $ab$.

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by $ab$, which is the LCD of the fractions. That will cancel all of the denominators.

Now we are ready to solve the following equations:

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

In this case we need to modify the equations before we can do elimination or substitution. And we only need to make a small change to arrange the equations for substitution: we just need to multiply both sides of the equation by $-\text{1}$ to remove the negative sign from $a$.

First rearrange things to set up the substitution:

Now do the substitution and complete the solution to find $b$.

Great - we have the first value, $b = 4$.

Now the last step: find the value of $a$. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

The answer is the pair of numbers $a = -10$ and $b = 4$. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

The values which solve these two equations are $a = -10$ and $b = 4$.

We need to solve the equations in this equation using elimination. (Substitution will also work, but the question says that we should use elimination.) As always for elimination, we need to use the two equations to cancel some of the terms.

By comparing the coefficients of the terms, we can see that the second $y$-term has a coefficient which is a multiple of the first $y$-term. In fact, $\text{6}$ is $\text{2}$ times as much as $\text{3}$. We can make both coefficients of $y$ the same by using this factor of $\text{2}$ to multiply the entire first equation:

Now we can eliminate the $y$-terms by subtracting one equation from the other. Then we can solve for $x$.

Now we can solve for $x$:

The last step is to substitute $y$ back into either of the equations so that we can find $x$. Here we will use the first equation:

The answers are $x = 3$ and $y = -4$.

We need to solve the equations in this equation using elimination. (Substitution will also work, but the question says that we should use elimination.) As always for elimination, we need to use the two equations to cancel some of the terms.

By comparing the coefficients of the terms, we can see that the second $x$-term has a coefficient which is a multiple of the first $x$-term. In fact, $\text{4}$ is $\text{2}$ times as much as $\text{2}$. We can make both coefficients of $x$ the same by using this factor of $\text{2}$ to multiply the entire first equation:

Now we can eliminate the $x$-terms by subtracting one equation from the other. Then we can solve for $y$.

Now we can solve for $y$:

The last step is to substitute $y$ back into either of the equations so that we can find $x$. Here we will use the first equation:

The answers are $x = 6$ and $y = 2$.

We need to solve the equations in this equation using elimination. (Substitution will also work, but the question says that we should use elimination.) As always for elimination, we need to use the two equations to cancel some of the terms.

By comparing the coefficients of the terms, we can see that the second $x$-term has a coefficient which is a multiple of the first $x$-term. In fact, $\text{10}$ is $\text{10}$ times as much as $\text{1}$. We can make both coefficients of $x$ the same by using this factor of $\text{10}$ to multiply the entire first equation:

Now we can eliminate the $x$-terms by subtracting one equation from the other. Then we can solve for $y$.

Now we can solve for $y$:

The last step is to substitute $y$ back into either of the equations so that we can find $x$. Here we will use the first equation:

The answers are $x = 0$ and $y = -1$.

The best choice for substitution is based on which variable is easiest to isolate. In this case, you should look for the variable which is already isolated.

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is substitution. The substitution method works by using one of the equations to replace one of the variables in the other equation. We need to figure out which variable is the best choice to substitute for these two equations.

In these two equations, there are two $x$-terms and two $y$-terms:

The goal of substitution is to use one of the equations to remove a variable from the other equation.

In this case, the best choice is to substitute the $y$ from the second equation into the first equation. This is because in the second equation the $y$ is already isolated. To use substitution, we need a variable which is isolated (alone on one side of the equation).

In this case, the substitution looks like this:

Combining these equations by substitution, we get:

Since the second equation tells us that $y$ is equal to $y=5 x + 3$, we can substitute it into the other equation in place of $y$ straight away.

The best choice is to substitute the $y$ from the second equation into the first equation because it is already isolated.

The best choice for substitution is based on which variable is easiest to isolate. In this case, you should look for the variable which is already isolated.

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is substitution. The substitution method works by using one of the equations to replace one of the variables in the other equation. We need to figure out which variable is the best choice to substitute for these two equations.

In these two equations, there are two $x$-terms and two $y$-terms:

The goal of substitution is to use one of the equations to remove a variable from the other equation.

In this case, the best choice is to substitute the $y$ from the second equation into the first equation. This is because in the second equation the $y$ is already isolated. To use substitution, we need a variable which is isolated (alone on one side of the equation).

In this case, the substitution looks like this:

Combining these equations by substitution, we get:

Since the second equation tells us that $y$ is equal to $y=- 3 x - 1$, we can substitute it into the other equation in place of $y$ straight away.

The best choice is to substitute the $y$ from the second equation into the first equation because it is already isolated.

The best choice for substitution is based on which variable is easiest to isolate. In this case, you should look for the variable which is already isolated.

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is substitution. The substitution method works by using one of the equations to replace one of the variables in the other equation. We need to figure out which variable is the best choice to substitute for these two equations.

In these two equations, there are two $x$-terms and two $y$-terms:

The goal of substitution is to use one of the equations to remove a variable from the other equation.

In this case, the best choice is to substitute the $x$ from the second equation into the first equation. This is because in the second equation the $x$ is already isolated. To use substitution, we need a variable which is isolated (alone on one side of the equation).

In this case, the substitution looks like this:

Combining these equations by substitution, we get:

Since the second equation tells us that $x$ is equal to $x=4 y - 5$, we can substitute it into the other equation in place of $x$ straight away.

The best choice is to substitute the $x$ from the second equation into the first equation because it is already isolated.

Remember that the goal of elimination is to cancel either the $k$-terms or the $p$-terms. So you need to change one of the coefficients in the second equations so that you can cancel terms.

For this question we need to multiply the second equation by a number so that we can solve the equations by elimination. In the original equations, none of the coefficients were ready to cancel (because none of them are equal). The first equation was already multiplied by $\text{3}$. We need to change one of the coefficients in the second equation so that we can cancel terms.

Specifically, we need to multiply $\text{2}$ to get $-\text{9}$ so we can cancel the $k$-terms, or multiply the $\text{3}$ to get $\text{15}$ so we can cancel the $p$-terms.

The numbers in these equations point us toward cancelling the $p$-terms. This is because $\text{15}$ is a multiple of $\text{3}$. So if we multiply the second equation by $\text{5}$ the the coefficients will be equal.

The coefficients we want to cancel are equal. So eliminating the terms requires subtracting both sides of the equation:

The correct answer can be either of these numbers: $\text{5}$ or $-\text{5}$.

Remember that the goal of elimination is to cancel either the $m$-terms or the $n$-terms. So you need to change one of the coefficients in the first equations so that you can cancel terms.

For this question we need to multiply the first equation by a number so that we can solve the equations by elimination. In the original equations, none of the coefficients were ready to cancel (because none of them are equal). The second equation was already multiplied by $\text{5}$. We need to change one of the coefficients in the first equation so that we can cancel terms.

Specifically, we need to multiply $\text{5}$ to get $\text{10}$ so we can cancel the $m$-terms, or multiply the $\text{4}$ to get $\text{15}$ so we can cancel the $n$-terms.

The numbers in these equations point us toward cancelling the $m$-terms. This is because $\text{10}$ is a multiple of $\text{5}$. So if we multiply the first equation by $\text{2}$ the the coefficients will be equal.

The coefficients we want to cancel are equal. So eliminating the terms requires subtracting both sides of the equation:

The correct answer can be either of these numbers: $\text{2}$ or $-\text{2}$.

Remember that the goal of elimination is to cancel either the $x$-terms or the $y$-terms. So you need to change one of the coefficients in the second equations so that you can cancel terms.

For this question we need to multiply the second equation by a number so that we can solve the equations by elimination. In the original equations, none of the coefficients were ready to cancel (because none of them are equal). The first equation was already multiplied by $\text{5}$. We need to change one of the coefficients in the second equation so that we can cancel terms.

Specifically, we need to multiply $\text{5}$ to get $-\text{15}$ so we can cancel the $x$-terms, or multiply the $\text{2}$ to get $\text{25}$ so we can cancel the $y$-terms.

The numbers in these equations point us toward cancelling the $x$-terms. This is because $-\text{15}$ is a multiple of $\text{5}$. So if we multiply the second equation by $\text{3}$ the the coefficients will be equal and opposite.

The coefficients we want to cancel are equal and opposite. So eliminating the terms requires adding both sides of the equation:

The correct answer can be either of these numbers: $-\text{3}$ or $\text{3}$.

You need to find the coordinates of the point where the lines intersect.

This question is about two equations. We need to find the values of $x$ and $y$ which solve the equations 'simultaneously'. This means we want one pair of $x$ and $y$ values which solves both equations.

There are different methods to solve equations simultaneously. In this case we can use the graph: the solutions we want are the coordinates where the lines intersect. The intersection point is very special - it is the only point shared by both lines. And the coordinates of that point are the only numbers which solve both equations! In this case the lines intersect at the point $\left(-1; 3\right)$.

We can prove that this is correct by substituting the values into each of the equations:

and

Perfect - the values $x = -1$ and $y = 3$ solve the equations simultaneously.

The correct answers are $x = -1$ and $y = 3$.

You need to find the coordinates of the point where the lines intersect.

This question is about two equations. We need to find the values of $x$ and $y$ which solve the equations 'simultaneously'. This means we want one pair of $x$ and $y$ values which solves both equations.

There are different methods to solve equations simultaneously. In this case we can use the graph: the solutions we want are the coordinates where the lines intersect. The intersection point is very special - it is the only point shared by both lines. And the coordinates of that point are the only numbers which solve both equations! In this case the lines intersect at the point $\left(-3; 2\right)$.

We can prove that this is correct by substituting the values into each of the equations:

and

Perfect - the values $x = -3$ and $y = 2$ solve the equations simultaneously.

The correct answers are $x = -3$ and $y = 2$.

You need to find the coordinates of the point where the lines intersect.

This question is about two equations. We need to find the values of $x$ and $y$ which solve the equations 'simultaneously'. This means we want one pair of $x$ and $y$ values which solves both equations.

There are different methods to solve equations simultaneously. In this case we can use the graph: the solutions we want are the coordinates where the lines intersect. The intersection point is very special - it is the only point shared by both lines. And the coordinates of that point are the only numbers which solve both equations! In this case the lines intersect at the point $\left(3; 4\right)$.

We can prove that this is correct by substituting the values into each of the equations:

and

Perfect - the values $x = 3$ and $y = 4$ solve the equations simultaneously.

The correct answers are $x = 3$ and $y = 4$.

The first thing you need to do (if you will use substitution to solve the pair of equations) is isolate one of the variables from one of the equations; then substitute what you get into the other equation.

When we solve simultaneous equations, we are working to find values for $x$ and $y$ which make both equations true!

We can solve these equations by substitution. Substitution requires isolating a variable. We need to look at the equations to decide which variable to isolate. We can see that the second equation looks good for this, because the $x$ has a coefficient of one:

Isolate $x$ in the second equation, which can be done in one step:

Now we can substitute the expression $- 5 y - 19$ into the first equation and solve:

Right now we have only half of the answer! To get the other half, we need to find the value of $x$. We can use the equation we got when we isolated $x$: it is the most convenient choice because $x$ is the subject of the equation.

The correct answers are $x = 6$ and $y = -5$.

The first thing you need to do (if you will use substitution to solve the pair of equations) is isolate one of the variables from one of the equations; then substitute what you get into the other equation.

When we solve simultaneous equations, we are working to find values for $x$ and $y$ which make both equations true!

We can solve these equations by substitution. Substitution requires isolating a variable. We need to look at the equations to decide which variable to isolate. We can see that the second equation looks good for this, because the $x$ has a coefficient of one:

Isolate $x$ in the second equation, which can be done in one step:

Now we can substitute the expression $- 4 y - 7$ into the first equation and solve:

Right now we have only half of the answer! To get the other half, we need to find the value of $x$. We can use the equation we got when we isolated $x$: it is the most convenient choice because $x$ is the subject of the equation.

The correct answers are $x = 9$ and $y = -4$.

The first thing you need to do (if you will use substitution to solve the pair of equations) is isolate one of the variables from one of the equations; then substitute what you get into the other equation.

When we solve simultaneous equations, we are working to find values for $x$ and $y$ which make both equations true!

We can solve these equations by substitution. Substitution requires isolating a variable. We need to look at the equations to decide which variable to isolate. We can see that the second equation looks good for this, because the $x$ has a coefficient of one:

Isolate $x$ in the second equation, which can be done in one step:

Now we can substitute the expression $6 y + 36$ into the first equation and solve:

Right now we have only half of the answer! To get the other half, we need to find the value of $x$. We can use the equation we got when we isolated $x$: it is the most convenient choice because $x$ is the subject of the equation.

The correct answers are $x = 6$ and $y = -5$.

You need to plot the graphs of the two equations on the same Cartesian plane.

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: $y = x^{2} + 3 x - 4$.

To plot this graph, we need to find the coordinates of the points where the graph will cross the $x$- and $y$-axes. These points are the $x$- and $y$-intercepts, respectively. We will then use these points to plot the graph.

The $y$-intercept is the quantity in the equation which is not multiplied by $x$. For this equation, the $y$-intercept is $\text{−4}$. The coordinates of this point are $\left(0; -4\right)$.

The $x$-intercepts are the roots of the equation $x^{2} + 3 x - 4 = 0$. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{3}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{4}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $-\text{4}$. The two numbers are $\text{4}$ and $-\text{1}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are $x = 1$ and $x = -4$. The coordinates of these roots are $\left(1; 0 \right)$ and $\left(-4; 0 \right)$.

Using these $x$- and $y$-intercepts we will draw the graph:

As we did for the quadratic equation, we will first find the $y$-intercept. After that we will calculate root for the linear equation. We remember that the $y$-intercept is the quantity in the equation which is not multiplied by $x$.

The $y$-intercept is $-\text{7}$. Its coordinates are: $\left(0; -7\right)$.

We will work out the root of the equation $0 = - x - 7$.

The root is: $x = -7$. It's coordinates are: $\left( -7; 0 \right)$.

We will now plot the equation $y = - x - 7$ on the same Cartesian plane we have used for the graph of the quadratic equation.

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: $(-1; -6)$ and $(-3; -4)$.

We have used a graphical method to determine the solutions of the simultaneous equations: $y = x^{2} + 3 x - 4$ and $y = - x - 7$. The solutions are the points where their graphs intersect.

The points of intersection are $(-1; -6)$ and $(-3; -4)$.

You need to plot the graphs of the two equations on the same Cartesian plane.

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: $y = x^{2} + 6 x + 8$.

To plot this graph, we need to find the coordinates of the points where the graph will cross the $x$- and $y$-axes. These points are the $x$- and $y$-intercepts, respectively. We will then use these points to plot the graph.

The $y$-intercept is the quantity in the equation which is not multiplied by $x$. For this equation, the $y$-intercept is $\text{8}$. The coordinates of this point are $\left(0; 8\right)$.

The $x$-intercepts are the roots of the equation $x^{2} + 6 x + 8 = 0$. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{6}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{8}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{8}$. The two numbers are $\text{2}$ and $\text{4}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are $x = -2$ and $x = -4$. The coordinates of these roots are $\left(-2; 0 \right)$ and $\left(-4; 0 \right)$.

Using these $x$- and $y$-intercepts we will draw the graph:

As we did for the quadratic equation, we will first find the $y$-intercept. After that we will calculate root for the linear equation. We remember that the $y$-intercept is the quantity in the equation which is not multiplied by $x$.

The $y$-intercept is $\text{2}$. Its coordinates are: $\left(0; 2\right)$.

We will work out the root of the equation $0 = - x + 2$.

The root is: $x = 2$. It's coordinates are: $\left( 2; 0 \right)$.

We will now plot the equation $y = - x + 2$ on the same Cartesian plane we have used for the graph of the quadratic equation.

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: $(-1; 3)$ and $(-6; 8)$.

We have used a graphical method to determine the solutions of the simultaneous equations: $y = x^{2} + 6 x + 8$ and $y = - x + 2$. The solutions are the points where their graphs intersect.

The points of intersection are $(-1; 3)$ and $(-6; 8)$.

You need to plot the graphs of the two equations on the same Cartesian plane.

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: $y = x^{2} - 7 x + 6$.

To plot this graph, we need to find the coordinates of the points where the graph will cross the $x$- and $y$-axes. These points are the $x$- and $y$-intercepts, respectively. We will then use these points to plot the graph.

The $y$-intercept is the quantity in the equation which is not multiplied by $x$. For this equation, the $y$-intercept is $\text{6}$. The coordinates of this point are $\left(0; 6\right)$.

The $x$-intercepts are the roots of the equation $x^{2} - 7 x + 6 = 0$. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{7}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{6}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{6}$. The two numbers are $-\text{1}$ and $-\text{6}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are $x = 6$ and $x = 1$. The coordinates of these roots are $\left(6; 0 \right)$ and $\left(1; 0 \right)$.

Using these $x$- and $y$-intercepts we will draw the graph:

As we did for the quadratic equation, we will first find the $y$-intercept. After that we will calculate root for the linear equation. We remember that the $y$-intercept is the quantity in the equation which is not multiplied by $x$.

The $y$-intercept is $-\text{1}$. Its coordinates are: $\left(0; -1\right)$.

We will work out the root of the equation $0 = x - 1$.

The root is: $x = 1$. It's coordinates are: $\left( 1; 0 \right)$.

We will now plot the equation $y = x - 1$ on the same Cartesian plane we have used for the graph of the quadratic equation.

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: $(7; 6)$ and $(1; 0)$.

We have used a graphical method to determine the solutions of the simultaneous equations: $y = x^{2} - 7 x + 6$ and $y = x - 1$. The solutions are the points where their graphs intersect.

The points of intersection are $(7; 6)$ and $(1; 0)$.

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate $x$.

Now we need to substitute the entire right side of the equation, $(- 8 y)$, into the quadratic equation in place of $x$. Be sure to use brackets! Then simplify the equation to get standard form.

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two $y$-values.

Cool! However, there is still work to do. We must now find the values of $x$ for these equations. We must substitute in the values $y = -3$ and $y = -1$. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally we have the answers: if $x = 24$ then $y = -3$ and if $x = 8$ then $y = -1$.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: $\left(24 ; -3\right)$ and $\left(8 ; -1\right)$.

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate $x$.

Now we need to substitute the entire right side of the equation, $(- 8 y + 3)$, into the quadratic equation in place of $x$. Be sure to use brackets! Then simplify the equation to get standard form.

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two $y$-values.

Cool! However, there is still work to do. We must now find the values of $x$ for these equations. We must substitute in the values $y = 4$ and $y = 0$. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally we have the answers: if $x = -29$ then $y = 4$ and if $x = 3$ then $y = 0$.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: $\left(-29 ; 4\right)$ and $\left(3 ; 0\right)$.

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate $x$.

Now we need to substitute the entire right side of the equation, $(- 2 y - 9)$, into the quadratic equation in place of $x$. Be sure to use brackets! Then simplify the equation to get standard form.

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two $y$-values.

Cool! However, there is still work to do. We must now find the values of $x$ for these equations. We must substitute in the values $y = -2$ and $y = 0$. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally we have the answers: if $x = -5$ then $y = -2$ and if $x = -9$ then $y = 0$.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: $\left(-5 ; -2\right)$ and $\left(-9 ; 0\right)$.

You need to start by substituting $2 x - 4$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{4}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{3}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{3}$. The two numbers are $-\text{1}$ and $-\text{3}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\text{3}$ and $\text{1}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = 3$:

For $x = 1$:

The solutions are $x = 3$ together with $y = 2$ , and $x = 1$ with $y = -2$. In coordinates pairs, the solutions are $\left(3; 2\right)$ and $\left(1; -2\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(3; 2\right)$ and $\left(1; -2\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(3; 2\right)$ and $\left(1; -2\right)$.

You need to start by substituting $x - 1$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{4}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{3}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{3}$. The two numbers are $-\text{1}$ and $-\text{3}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\text{3}$ and $\text{1}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = 3$:

For $x = 1$:

The solutions are $x = 3$ together with $y = 2$ , and $x = 1$ with $y = 0$. In coordinates pairs, the solutions are $\left(3; 2\right)$ and $\left(1; 0\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(3; 2\right)$ and $\left(1; 0\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(3; 2\right)$ and $\left(1; 0\right)$.

You need to start by substituting $2 x - 1$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{1}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{2}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $-\text{2}$. The two numbers are $\text{1}$ and $-\text{2}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\text{2}$ and $-\text{1}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = 2$:

For $x = -1$:

The solutions are $x = 2$ together with $y = 3$ , and $x = -1$ with $y = -3$. In coordinates pairs, the solutions are $\left(2; 3\right)$ and $\left(-1; -3\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(2; 3\right)$ and $\left(-1; -3\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(2; 3\right)$ and $\left(-1; -3\right)$.

You need to start by substituting $- 2 x - 3$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{11}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{18}$. This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $\text{6}$. The two numbers are $\text{2}$ and $\text{9}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $-\dfrac{2}{3}$ and $-\text{3}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = -\dfrac{2}{3}$:

For $x = -3$:

The solutions are $x = -\dfrac{2}{3}$ together with $y = -\dfrac{5}{3}$ , and $x = -3$ with $y = 3$. In coordinate pairs, the solutions are $\left(-\dfrac{2}{3}; -\dfrac{5}{3}\right)$ and $\left(-3; 3\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-\dfrac{2}{3}; -\dfrac{5}{3}\right)$ and $\left(-3; 3\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(-\dfrac{2}{3}; -\dfrac{5}{3}\right)$ and $\left(-3; 3\right)$.

You need to start by substituting $2 x + 3$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{5}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{6}$. This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $-\text{2}$. The two numbers are $\text{6}$ and $-\text{1}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\dfrac{1}{3}$ and $-\text{2}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = \dfrac{1}{3}$:

For $x = -2$:

The solutions are $x = \dfrac{1}{3}$ together with $y = \dfrac{11}{3}$ , and $x = -2$ with $y = -1$. In coordinate pairs, the solutions are $\left(\dfrac{1}{3}; \dfrac{11}{3}\right)$ and $\left(-2; -1\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(\dfrac{1}{3}; \dfrac{11}{3}\right)$ and $\left(-2; -1\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(\dfrac{1}{3}; \dfrac{11}{3}\right)$ and $\left(-2; -1\right)$.

You need to start by substituting $x - 1$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{9}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{14}$. This is the product of the coefficient of the first term $\text{2}$ and the value of the last term $\text{7}$. The two numbers are $-\text{2}$ and $-\text{7}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\dfrac{7}{2}$ and $\text{1}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = \dfrac{7}{2}$:

For $x = 1$:

The solutions are $x = \dfrac{7}{2}$ together with $y = \dfrac{5}{2}$ , and $x = 1$ with $y = 0$. In coordinate pairs, the solutions are $\left(\dfrac{7}{2}; \dfrac{5}{2}\right)$ and $\left(1; 0\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(\dfrac{7}{2}; \dfrac{5}{2}\right)$ and $\left(1; 0\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(\dfrac{7}{2}; \dfrac{5}{2}\right)$ and $\left(1; 0\right)$.

You need to start by substituting $y + 2$ in place of $x$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $x$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{1}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{2}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $-\text{2}$. The two numbers are $\text{2}$ and $-\text{1}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are 1 and -2.

We will substitute each $y$-value into the linear equation. This will give us the corresponding values of $x$.

For $y = 1$:

For $y = -2$:

The solutions are $x = 3$ together with $y = 1$ , and $x = 0$ with $y = -2$. In coordinates pairs, the solutions are $\left(3; 1\right)$ and $\left(0; -2\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(3; 1\right)$ and $\left(0; -2\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(3; 1\right)$ and $\left(0; -2\right)$.

You need to start by substituting $y + 1$ in place of $x$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $x$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{1}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{2}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $-\text{2}$. The two numbers are $\text{1}$ and $-\text{2}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are 2 and -1.

We will substitute each $y$-value into the linear equation. This will give us the corresponding values of $x$.

For $y = 2$:

For $y = -1$:

The solutions are $x = 3$ together with $y = 2$ , and $x = 0$ with $y = -1$. In coordinates pairs, the solutions are $\left(3; 2\right)$ and $\left(0; -1\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(3; 2\right)$ and $\left(0; -1\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(3; 2\right)$ and $\left(0; -1\right)$.

You need to start by substituting $y - 4$ in place of $x$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $x$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{6}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{8}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{8}$. The two numbers are $-\text{2}$ and $-\text{4}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are 4 and 2.

We will substitute each $y$-value into the linear equation. This will give us the corresponding values of $x$.

For $y = 4$:

For $y = 2$:

The solutions are $x = 0$ together with $y = 4$ , and $x = -2$ with $y = 2$. In coordinates pairs, the solutions are $\left(0; 4\right)$ and $\left(-2; 2\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(0; 4\right)$ and $\left(-2; 2\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(0; 4\right)$ and $\left(-2; 2\right)$.

Solve the equations simultaneously to find their intersection points.

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation $y$ is equal to the quantity $- \frac{9 x}{2} + \frac{15}{2}$; and at the same time, the quadratic equation also contains the value $y$. Substitute like this:

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is $\text{2}$. So multiply both sides of the equation by $\text{2}$ in order to cancel the denominators. Then arrange the equation in standard form.

In this case, there is a common factor of $\text{3}$ for all of the terms. Divide out this factor before factorising.

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two $x$-values.

Great... but it isn't over yet. We must now find the values of $y$ for these equations. Substitute in the values $x = -3$ and $x = 3$ into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally! The solutions are $\left(-3 ; 21\right)$ and $\left(3 ; -6\right)$.

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are $\left(-3 ; 21\right)$ and $\left(3 ; -6\right)$.

Solve the equations simultaneously to find their intersection points.

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation $y$ is equal to the quantity $\frac{3 x}{4} + \frac{21}{4}$; and at the same time, the quadratic equation also contains the value $y$. Substitute like this:

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is $\text{4}$. So multiply both sides of the equation by $\text{4}$ in order to cancel the denominators. Then arrange the equation in standard form.

In this case, there is a common factor of $\text{3}$ for all of the terms. Divide out this factor before factorising.

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two $x$-values.

Great... but it isn't over yet. We must now find the values of $y$ for these equations. Substitute in the values $x = -5$ and $x = -1$ into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally! The solutions are $\left(-5 ; \frac{3}{2}\right)$ and $\left(-1 ; \frac{9}{2}\right)$.

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are $\left(-5 ; \frac{3}{2}\right)$ and $\left(-1 ; \frac{9}{2}\right)$.

Solve the equations simultaneously to find their intersection points.

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation $y$ is equal to the quantity $- \frac{x}{3} - 1$; and at the same time, the quadratic equation also contains the value $y$. Substitute like this:

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is $\text{3}$. So multiply both sides of the equation by $\text{3}$ in order to cancel the denominators. Then arrange the equation in standard form.

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two $x$-values.

Great... but it isn't over yet. We must now find the values of $y$ for these equations. Substitute in the values $x = -3$ and $x = 2$ into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally! The solutions are $\left(-3 ; 0\right)$ and $\left(2 ; - \frac{5}{3}\right)$.

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are $\left(-3 ; 0\right)$ and $\left(2 ; - \frac{5}{3}\right)$.

You need to substitute $- x + 5$ in place of $y$ in the quadratic equation $y = x^{2} + x - 3$.

We have two equations that we will solve by substitution. The equations are:

We need to substitute $- x + 5$ from the linear equation in place of $y$ in the quadratic equation.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for $x$. The question does not ask us to solve the simultaneous equations.

The correct substitution step is $- x + 5=x^{2} + x - 3$ which is choice A.

You need to substitute $- x + 2$ in place of $y$ in the quadratic equation $y = x^{2} - 2 x - 4$.

We have two equations that we will solve by substitution. The equations are:

We need to substitute $- x + 2$ from the linear equation in place of $y$ in the quadratic equation.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for $x$. The question does not ask us to solve the simultaneous equations.

The correct substitution step is $- x + 2=x^{2} - 2 x - 4$ which is choice C.

You need to substitute $y + 5$ in place of $x$ in the quadratic equation $y = x^{2} - 4 x + 1$.

We have two equations that we will solve by substitution. The equations are:

We need to substitute $y + 5$ from the linear equation in place of $x$ in the quadratic equation.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for $y$. The question does not ask us to solve the simultaneous equations.

The correct substitution step is $y=(y + 5)^2-4(y + 5)+1$ which is choice D.

You need to start by substituting $- 2 x - 5$ in place of $y$ in the quadratic equation and simplify. You must expect a special kind of solution.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is $y = 3 x^{2} + 2 x + 1$ and the linear equation is $y = - 2 x - 5$. This means we expect two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We need two numbers that add to give $\text{4}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{18}$. This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $\text{6}$. In this case, there are no such numbers. We will now use the quadratic formula to solve this equation.

The solution for a quadratic equation of the form $ax^2 + bx + c = 0$:

Before we can use the quadratic formula, we need to identify the coefficients $a$, $b$ and $c$ from the equation $3 x^{2}+4 x+6 = 0$ These are:

We will substitute these values into the formula and simplify.

The number under the square root sign is negative. This means that our quadratic equation has no real solution. This means that $y = 3 x^{2} + 2 x + 1$ and $y = - 2 x - 5$ cannot be solved simultaneously.

We can confirm our result by plotting the graph of each equation on the same set of axes. We do not expect these graphs to intersect.

Aha! The two graphs do not have an intersection point. This means that indeed the two equations have no solution.

The final answer is no solution.

You need to start by substituting $3 x - 2$ in place of $y$ in the quadratic equation and simplify. You must expect a special kind of solution.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is $y = x^{2} - x + 2$ and the linear equation is $y = 3 x - 2$. This means we expect two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We will factorise using the grouping method. For this method, we need two numbers that add to give $-\text{4}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{4}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{4}$. If we take $-\text{2}$ which is half of the value of $-\text{4}$ (the coefficient of the middle term) and square it, we get the product $\text{4}$. In this case we will use only one number to factorise, that is, $-\text{2}$. We will re-write the middle term, forming two terms, using this value.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, $\text{2}$. This root is the $x$-value of our solution. This $x$-value will have one corresponding $y$ value.

Now that we have the value of $x$, we will substitute it into the linear equation to calculate the corresponding value of $y$.

The solution is $x = 2$ together with $y = 4$. Using coordinate pairs, the solution is $\left(2; 4\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(2; 4\right)$. This is the same solution we have calculated. This means we are $\text{100}\%$ correct.

The solution is $\left(2; 4\right)$.

You need to start by substituting $- 2 x + 1$ in place of $y$ in the quadratic equation and simplify. You must expect a special kind of solution.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is $y = x^{2} + 2$ and the linear equation is $y = - 2 x + 1$. This means we expect two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We will factorise using the grouping method. For this method, we need two numbers that add to give $\text{2}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{1}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{1}$. If we take $\text{1}$ which is half of the value of $\text{2}$ (the coefficient of the middle term) and square it, we get the product $\text{1}$. In this case we will use only one number to factorise, that is, $\text{1}$. We will re-write the middle term, forming two terms, using this value.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, $-\text{1}$. This root is the $x$-value of our solution. This $x$-value will have one corresponding $y$ value.

Now that we have the value of $x$, we will substitute it into the linear equation to calculate the corresponding value of $y$.

The solution is $x = -1$ together with $y = 3$. Using coordinate pairs, the solution is $\left(-1; 3\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-1; 3\right)$. This is the same solution we have calculated. This means we are $\text{100}\%$ correct.

The solution is $\left(-1; 3\right)$.

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.

We need to solve the following pair of equations for $x$ and $y$:

First, we can substitute $- 3 y$ in place of $x$ in the second equation:

We can simplify this equation to solve for $y$:

$y$ can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about $y^{2}$. So we have to consider both the positive and the negative values as solutions to the equations.

Each of the $y$-values that we have found will have a corresponding value of $x$, which can be found from the equation $x = - 3 y$.

For $y = 2$:

And for $y = -2$:

So the $x$-values are $x = -6 \text{ or } x = 6$.

Therefore the correct coordinate pairs are $(-6; 2)$ and $(6; -2)$.

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.

We need to solve the following pair of equations for $x$ and $y$:

First, we can substitute $2 y$ in place of $x$ in the second equation:

We can simplify this equation to solve for $y$:

$y$ can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about $y^{2}$. So we have to consider both the positive and the negative values as solutions to the equations.

Each of the $y$-values that we have found will have a corresponding value of $x$, which can be found from the equation $x = 2 y$.

For $y = 2$:

And for $y = -2$:

So the $x$-values are $x = 4 \text{ or } x = -4$.

Therefore the correct coordinate pairs are $(4; 2)$ and $(-4; -2)$.

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.

We need to solve the following pair of equations for $x$ and $y$:

First, we can substitute $2 y$ in place of $x$ in the second equation:

We can simplify this equation to solve for $y$:

$y$ can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about $y^{2}$. So we have to consider both the positive and the negative values as solutions to the equations.

Each of the $y$-values that we have found will have a corresponding value of $x$, which can be found from the equation $x = 2 y$.

For $y = 4$:

And for $y = -4$:

So the $x$-values are $x = 8 \text{ or } x = -8$.

Therefore the correct coordinate pairs are $(8; 4)$ and $(-8; -4)$.

You need to start by substituting $- y + 3$ in place of $x$ in the hyperbolic equation and then solve for $y$.

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $x$ from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are $\text{3}$ and $\text{6}$.

By using the values of $y$, we will calculate the corresponding values of $x$. We will do so by substituting each $y$-value into the linear equation.

For $y = 3$:

For $y = 6$:

The solutions are $x = 0$ together with $y = 3$ , and $x = -3$ with $y = 6$. Using coordinate pairs, the solutions are $\left(0; 3\right)$ and $\left(-3; 6\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(0; 3\right)$ and $\left(-3; 6\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(0; 3\right)$ and $\left(-3; 6\right)$.

You need to start by substituting $- y - 1$ in place of $x$ in the hyperbolic equation and then solve for $y$.

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $x$ from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are $-\text{5}$ and $-\text{1}$.

By using the values of $y$, we will calculate the corresponding values of $x$. We will do so by substituting each $y$-value into the linear equation.

For $y = -5$:

For $y = -1$:

The solutions are $x = 4$ together with $y = -5$ , and $x = 0$ with $y = -1$. Using coordinate pairs, the solutions are $\left(4; -5\right)$ and $\left(0; -1\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(4; -5\right)$ and $\left(0; -1\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(4; -5\right)$ and $\left(0; -1\right)$.

You need to start by substituting $- 2 x - 3$ in place of $y$ in the hyperbolic equation and then solve for $x$.

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $y$ from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $-\text{1}$ and $-\text{5}$.

By using the values of $x$, we will calculate the corresponding values of $y$. We will do so by substituting each $x$-value into the linear equation.

For $x = -1$:

For $x = -5$:

The solutions are $x = -1$ together with $y = -1$ , and $x = -5$ with $y = 7$. Using coordinate pairs, the solutions are $\left(-1; -1\right)$ and $\left(-5; 7\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-1; -1\right)$ and $\left(-5; 7\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(-1; -1\right)$ and $\left(-5; 7\right)$.